<h3>The time in fourth race is 43.04 seconds</h3>
<em><u>Solution:</u></em>
Given that,
Shelly is trying to improve her running time for a track race
She ran the first race in 43.13 seconds
Her time was 43.1 seconds in the second race and 43.07 seconds in the third race
<em><u>Therefore, pattern is:</u></em>
43.13, 43.1, 43.07 , ..
<em><u>Find the difference between each terms</u></em>
d = 43.13 - 43.1 = 0.03
d = 43.1 - 43.07 = 0.03
Thus the difference is same
<em><u>If this pattern continues what will shellys time be in the fourth race?</u></em>
Fourth race = third race - 0.03
Fourth race = 43.07 - 0.03 = 43.04
Thus the time in fourth race is 43.04 seconds
Answer:
1/50000
Step-by-step explanation:
Sorry if this wasn't helpful! this was the only logical answer i had!
Therefore the answer is B.
Hope this helps !
Photon
Answer:
The inequality tha can be used to find how many more bags of popcorn Jeff still needs to sell today to make a profit is
x - 12 ≥ 40.
Step-by-step explanation:
Let us represent the number of bags of popcorn as p
Jeff sells popcorn for $3 per bag. To make a profit each day, he needs to sell at least $120 worth of popcorn.
We have the equation:
$3 × p ≥ $120
3p ≥ $120
p ≥ 120/3
p ≥ 40 bags
He must sell at least 40 bags to make a profit
He has sold $36 worth of popcorn today.
1 bag is $3
Hence, he has sold
$36/$3 = 12 bags of popcorn
Which inequality can be used to find Jeff still needs to sell today to make a profit?
This equality is given as:
x - 12 bags ≥ 40 bags
x - 12 ≥ 40
x ≥40 - 12
x ≥ 28 bags
She still need to sell 28 more bags of popcorn.
Answer: The proof is mentioned below.
Step-by-step explanation:
Here, Δ ABC is isosceles triangle.
Therefore, AB = BC
Prove: Δ ABO ≅ Δ ACO
In Δ ABO and Δ ACO,
∠ BAO ≅ ∠ CAO ( AO bisects ∠ BAC )
∠ AOB ≅ ∠ AOC ( AO is perpendicular to BC )
BO ≅ OC ( O is the mid point of BC)
Thus, By ASA postulate of congruence,
Δ ABO ≅ Δ ACO
Therefore, By CPCTC,
∠B ≅ ∠ C
Where ∠ B and ∠ C are the base angles of Δ ABC.
